Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Similar presentations

Presentation on theme: "Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal."— Presentation transcript:

5
1. Uniform Motion Type of circular motion which occurs when an object moves at constant speed in a circular path. Circular motion which Object’s speed around its path remains constant with changing velocity or changing direction

9
5.Critical Point – CRITICAL VELOCITY Critical point occurs at the highest point of the circular motion. The velocity at the highest point of a vertical uniform circular motion is Critical Velocity. Critical point or critical velocity happens when the tension on the string is 0 N. This is the speed necessary to maintain circular motion.

15
CW: 1.A 50 cm rope is tied to the handle of the bucket which is then whirled in a vertical circle. The mass of the bucket is 2 kg. a.What is the speed of the bucket if the tension of the rope at the lowest point of its path is 40 N ? b. What is the critical speed if the rope would become slack when the bucket reaches the highest point in the circle ?

16
2. An object moves at a constant speed in a circular path of radius r at a rate of 2 revolution per second. What is its acceleration in terms of r and π ? 3. An object m =2 kg at the end of the string is whirled around in the vertical circle. The circular motion has a radius of 0.5 m. If the speed of the object is 4m/s at the bottom of the circle a. What is the tension of the string at that point ? b. What is the minimum speed necessary for the object to obtain circular motion ?

20
Experiment : Roller Coaster PURPOSE To allow the marble to complete the 1 and 2 loop track from start to finish. To calculate the speed of the marble form start to finish on 1 and 2 loop track To calculate the potential energy and the kinetic energy of the marble for each track To calculate the centripetal acceleration and centripetal force for 1 loop and 2 loop track.

25
ANALYSIS AND CONCLUSIONS COMPARE THE 1 LOOP AND 2 LOOP TRACK ROLLER COASTER COMPARE YOUR ROLLER COASTER WITH 2 OTHER GROUPs’ ROLER COASTER WITH DIFFERENT STARTING HEIGHT. COMPARE THEIR RESULTS AND YOURS. HOW DOES THE START HEIGHT AFFECTS ALL OF THE VARIABLES ?

34
Solving Torque Problems at Equilibrium 1.Identify all Forces along x 2.Identify all Forces along y 3.Set up the equation EFx = 0 at equilibrium 4.Set up the equation EFy = 0 at equilibrium 5.Choose a pivot point to set up equation for ET= 0 6.Torque at the pivot point is zero. 7.Torque = perpendicular Force X lever arm 8.Clockwise Torque is negative 9.Counterclockwise Torque is positive.

35
Examples on Torque 24. A uniform bar of mass m and the length L extends horizontally from a wall. A supporting wire connects the wall to the bar’s midpoint, making an angle of 55 o with the bar. A sign of mass M hangs from the end of the bar. If the system is in static equilibrium and the wall has friction, determine the tension in the wire and the strength of the force exerted on the bar by the wall if M =8kg and m=2 kg. 55 o m M

40
C W : TORQUE 1. A solid cylinder consisting of an outer radius R1 and an inner radius R2 is pivoted on a frictionless axle as shown below. A string is wound around the outer radius and is pulled to the right with the force F1=3N. A second string is wound around the inner radius and is pulled down with the force F2=5N. If R1 = 0.75 m and R2= 0.35m, what is the net torque acting on the cylinder? F1 =3N F2= 5N

41
2. In an effort to tighten a bolt, a force F is applied as shown in the figure above. If the distance from the end of the wrench to the center of the bolt is 20 cm and F = 20 N, What is the magnitude of the torque produced by F? F

42
3. A uniform meter stick of mass 1 kg is hanging from a thread attached at the stick’s midpoint. One block of mass m =3kg hangs from the left end of the stick and another block, of unknown mass M, hangs below the 80 cm mark on the meter stick. If the stick remains at rest in the horizontal position shown above, what is M ? m M

44
Newton’s Law of Gravitation 27. gravitation force -Any two objects in the universe exert an attractive force on each other 28. G – is the Universal Gravitational Constant = 6.67 X Nm 2 /kg Newton’s Law of Gravitation : Any two objects in the universe exert an attractive force on each other called gravitational force whose strength is proportional to the product of the object’s masses and inversely proportional to the square of the distance between them as measured from center to center. 30. F G = Gm 1 m 2 / r 2

47
33. Communications satellites are often parked in geosynchoronous orbits above the Earth’s surface. These satellites have orbit periods that are equal to earth’s rotation period, so they remain above the same position on Earth’s surface. Determine the altitude that a satellite must have to be in geosynchoronous orbit above a fixed position on earth’s equator. ( the mass of the earth is 5.98 X kg ) Let m – satellite’s mass M- the mass of the earth R – distance between the center of the earth and the center of the satellite F G = F c = mv 2 / r